
Re: Inverting elliptic integrals
Posted:
Aug 9, 2004 3:08 PM


mathar@mpia.de wrote: > In article <muqqag9mwbv2@legacy>, > David Powers <powers@computer.org> writes: > >I need the inverse of the (incomplete) elliptic integral of the second > >kind. It is well defined but I can't track down a solution or code > >for it. > >.... > > For a numerical representation of the inverse in terms of the angle phi, > where E(phi,m)=int(theta=0..phi) sqrt(1m*sin^2 theta) dtheta is the > elliptic integral fo the second kind, one could expand E(phi,m) in a > power series around phi=0, > > Emphi := phi > 1/6*m*phi^3 > +1/5*(1/6*m1/8*m^2)*phi^5 > +1/7*(1/45*m+1/12*m^21/16*m^3)*phi^7 > +1/9*(1/630*m1/40*m^2+1/16*m^35/128*m^4)*phi^9 ... > > The expansion coefficients in front of the order phi^(n+1) (n=2,4,6,...) > are [sum over k=2,4,..,n of U(k,n)*F(k,m)/k!]/(n+1)! > where > U(k,n) = (1)^[(k+n)/2]/2^(kn)*[sum l=0,1,...k of (1)^l > (lk/2)^n*binom(k,l)] and > F(k,m) = m^(n/2)*[(k1)!!]^2/(k1) , > with (k1)!! = 1*3*5*7*...*(k1) . > > Then invert this as outlined in chapt 3.6.25 of the book edited by M > Abramowitz and I Stegun: > > phi := E(phi,m) > +1/6*m*E(phi,m)^3 > +1/120*m*(13*m4)*E(phi,m)^5 > +1/5040*m*(493*m^2284*m+16)*E(phi,m)^7 > +1/362880*m*(37369*m^331224*m^2+4944*m64)*E(phi,m)^9 ...
Yes. That's the same series I posted last week. See my Out[2], in which z is the same as your E(phi,m).
Of course, the coefficients of the polynomials in m are obtainable by reversion of series, as we both did. But does anyone know of a simpler way of calculating them? (I may submit the sequence of coefficients to the OEIS in the near future.)
Regards, David Cantrell

